Решебник по геометрии 9 класс Мерзляк Задание 35

\[Схематический\ рисунок.\]

\[Дано:\]

\[ABCD - параллелограмм;\]

\[AB = 2\sqrt{2}\ см;\]

\[AD = 5\ см;\]

\[\angle A = \angle C = 45{^\circ}.\]

\[Найти:\]

\[AC;\ BD.\]

\[Решение.\]

\[1)\ В\ \mathrm{\Delta}ABD:\]

\[BD^{2} =\]

\[= AB^{2} + AD^{2} - 2AB \bullet AD \bullet \cos{\angle A};\]

\[= 8 + 25 - 2 \bullet 2\sqrt{2} \bullet 5 \bullet \cos{45{^\circ}} =\]

\[= 33 - 20\sqrt{2} \bullet \frac{\sqrt{2}}{2} =\]

\[= 33 - 20 = 13.\]

\[BD = \sqrt{13}\ см.\]

\[2)\ В\ ABCD:\]

\[\angle B = 180{^\circ} - \angle A = 135{^\circ};\]

\[BC = AD = 5\ см.\]

\[3)\ В\ \mathrm{\Delta}ABC:\]

\[AC^{2} =\]

\[= AB^{2} + BC^{2} - 2AB \bullet BC \bullet \cos{\angle B} =\]

\[= 8 + 25 - 2 \bullet 2\sqrt{2} \bullet 5 \bullet \cos{135{^\circ}} =\]

\[= 33 + 20\sqrt{2} \bullet \frac{\sqrt{2}}{2} =\]

\[= 33 + 20 = 53.\]

\[AC = \sqrt{53}.\]

\[Ответ:\ \ \sqrt{13}\ см;\ \sqrt{53}\ см.\]